Problem: Find the distance between the point ${(-2, 8)}$ and the line $\enspace {y = \dfrac{1}{3}x - 8}\thinspace$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
First, find the equation of the perpendicular line that passes through ${(-2, 8)}$ The slope of the blue line is ${\dfrac{1}{3}}$ , and its negative reciprocal is ${-3}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = -3x + b}\thinspace$ We can plug our point, ${(-2, 8)}$ , into this equation to solve for ${b}$ , the y-intercept. $8 = {-3}(-2) + {b}$ $8 = 6 + {b}$ $8 - 6 = {b} = 2$ The equation of the perpendicular line is $\enspace {y = -3x + 2}\thinspace$ We can see from the graph (or by setting the equations equal to one another) that the two lines intersect at the point ${(3, -7)}$ . Thus, the distance we're looking for is the distance between the two red points. The distance formula tells us that the distance between two points is equal to: $\sqrt{( x_{1} - x_{2} )^2 + ( y_{1} - y_{2} )^2}$ Plugging in our points ${(-2, 8)}$ and ${(3, -7)}$ gives us: $\sqrt{( {-2} - {3} )^2 + ( {8} - {-7} )^2}$ $= \sqrt{( -5 )^2 + ( 15 )^2} = \sqrt{250} = 5\sqrt{10}$ The distance between the point ${(-2, 8)}$ and the line $\thinspace {y = \dfrac{1}{3}x - 8}\enspace$ is $\thinspace5\sqrt{10}$.